3x(4x-3)=-8x+6

Solution for 3x(4x-3)=-8x+6 equation:

3x(4x-3)=-8x+6

We move all terms to the left:

3x(4x-3)-(-8x+6)=0

We multiply parentheses

12x^2-9x-(-8x+6)=0

We get rid of parentheses

12x^2-9x+8x-6=0

We add all the numbers together, and all the variables

12x^2-1x-6=0

a = 12; b = -1; c = -6;
Δ = b2-4ac
Δ = -12-4·12·(-6)
Δ = 289
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{289}=17$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-17}{2*12}=\frac{-16}{24}
=-2/3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+17}{2*12}=\frac{18}{24}
=3/4
$